Question

$\int \left[\log (1+\cos x)-x\tan \left(\frac{x}{2}\right)\right]dx$

• A. $x\log (1+\tan x)$
• B. $x\log (1+\sin x)$
• C. $\log (1+\sin x)$
• D. $x\log (1+\cos x)$

Answer 1

Answer: (D)

$x\log (1+\cos x)$

$I=\int \log (1+\cos x)dx-\int x\tan \left(x/2\right)dx$
$I=x\log (1+\cos x)+\int x\frac{\sin x}{1+\cos x}dx-\int x\tan \frac{x}{2}$.
$=x\log (1+\cos x)$ second term gets cancel $[\because \frac{\sin x}{1+\cos x}=\tan \frac{x}{2}]$
$=x\log (1+\cos x)$